Point $P$ is located inside triangle $ABC$ so that angles $PAB, PBC,$ and $PCA$ are all congruent. The sides of the triangle have lengths $AB=13, BC=14,$ and $CA=15.$  Find $\tan \angle PAB.$
Answer: Let $\omega = \angle PAB = \angle PBC = \angle PCA,$ and let $x = AP,$ $y = BP,$ and $z = CP.$

[asy]
unitsize(0.4 cm);

pair A, B, C, P;
int a, b, c;

a = 14;
b = 15;
c = 13;

A = (5,12);
B = (0,0);
C = (14,0);
P = (c^2*a^2*A + a^2*b^2*B + b^2*c^2*C)/(c^2*a^2 + a^2*b^2 + b^2*c^2);

draw(anglemark(B,A,P,40),red); 
draw(anglemark(C,B,P,40),red); 
draw(anglemark(A,C,P,40),red); 
draw(A--B--C--cycle);
draw(A--P);
draw(B--P);
draw(C--P);

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$P$", P, NE);
label("$13$", (A + B)/2, NW);
label("$14$", (B + C)/2, S);
label("$15$", (A + C)/2, NE);

label("$x$", (A + P)/2, W);
label("$y$", (B + P)/2, NW);
label("$z$", (C + P)/2, NE);
[/asy]

Then by the Law of Cosines applied to triangles $ABP,$ $BCP,$ $CAP,$ we get
\begin{align*}
y^2 &= x^2 + 169 - 26x \cos \omega, \\
z^2 &= y^2 + 196 - 28y \cos \omega, \\
x^2 &= z^2 + 225 - 30z \cos \omega.
\end{align*}Adding these, we get $x^2 + y^2 + z^2 = x^2 + y^2 + z^2 + 590 - (26x + 28y + 30z) \cos \omega.$  Then
\[(26x + 28y + 30z) \cos \omega = 590,\]or $(13x + 14y + 15z) \cos \omega = 295.$

Also, $[ABP] + [BCP] + [CAP] = [ABC].$  By Heron's formula, $[ABC] = 84,$ so
\[\frac{1}{2} \cdot 13 \cdot x \sin \omega + \frac{1}{2} \cdot 14 \cdot y \sin \omega + \frac{1}{2} \cdot 15 \cdot z \sin \omega = 84.\]Then $(13x + 14y + 15z) \sin \omega = 168.$

Dividing the equations $(13x + 14y + 15z) \sin \omega = 168$ and $(13x + 14y + 15z) \cos \omega = 295,$ we get $\tan \omega = \boxed{\frac{168}{295}}.$